poj 3181 Dollar Dayz

Dollar Dayz

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8672		Accepted: 3233

Description

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, $2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are: 

        1 @ US$3 + 1 @ US$2

        1 @ US$3 + 2 @ US$1

        1 @ US$2 + 3 @ US$1

        2 @ US$2 + 1 @ US$1

        5 @ US$1

Write a program than will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of $1..$K (1 <= K <= 100).

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

5 3

Sample Output

5

Source

USACO 2006 January Silver

 

题意：用n元去兑换价值不超过k的物品，求有多少兑换方法

思路:dp[i][j]:用前i种价格配出价值j的方案数

则dp[i][j]=dp[i-1][j]+dp[i-1][j-i]+...+dp[i-1][j-k*i];其中k=floor(j/i)

这样直接做还是比较慢的，可以继续化简：

当上式中的j=j-i时

dp[i][j-i]=dp[i-1][j-i]+...+dp[i-1][j-k*i];

则dp[i][j]=dp[i-1][j]+dp[i][j-i];

这样一来就可以少用一层循环了

还要注意数据过大，long long存不下,可以采用高低位存储数大数

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
typedef unsigned long long ll;
#define N_MAX 1000+20
#define K_MAX 100+20
#define MOD 100000000000000000
int n, k;
ll dp[K_MAX][N_MAX][2];
int main() {
    scanf("%d%d",&n,&k);
    dp[0][0][1] = 1;
    for (int i = 1; i <= k;i++) {
        dp[i][0][1] = 1;
        for (int j = 1; j <= n;j++) {
            if (j >= i) {
                dp[i][j][0] = dp[i - 1][j][0] + dp[i][j - i][0];
                dp[i][j][1] = dp[i - 1][j][1] + dp[i][j - i][1];
                dp[i][j][0] += dp[i][j][1] / MOD;
                dp[i][j][1] %= MOD;
            }
            else {
                dp[i][j][0] = dp[i - 1][j][0];
                dp[i][j][1] = dp[i - 1][j][1];
            }
        }
    }
    if (dp[k][n][0]) {
        cout << dp[k][n][0];
    }
    cout << dp[k][n][1]<<endl;
    return 0;
}